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#1 Re: Help Me ! » confusing trig problem... » 2006-01-11 20:24:43

seiya_001

thank you for the information... big_smile
god bless..up

#2 Re: Help Me ! » confusing trig problem... » 2006-01-11 15:19:53

seiya_001

thanks a lot bro up
btw, do you have any reference where i can get a sumary of trig identity? thanks again... big_smile

#3 Re: Help Me ! » confusing trig problem... » 2006-01-11 12:39:15

seiya_001

phi as in 180 deg... or is it pi dunno

#4 Help Me ! » confusing trig problem... » 2006-01-11 03:29:27

seiya_001
Replies: 6

can anybody tell me how to change

1     phi
-    ( ∫ A sin x . cos nx dx)
phi    0

into:

A     1     phi
-   .  -   . ∫(sin (1+n)x + sin (1-n)x )dx
phi    2    0

thanks before... smile

#5 Re: Help Me ! » basic geometry... » 2005-10-24 13:25:54

seiya_001

thank you..thank you..thank you..thank you..thank you..thank you..thank you.. !! (btw i've corrected  the foolish mistake in my example...*grin*)

#6 Re: Help Me ! » basic geometry... » 2005-10-23 19:14:19

seiya_001

well it's one of them (for straight line only..) what about parabolic, hiperbolic, exponential, and so on..? but thanks anyway.. smile

#7 Help Me ! » basic geometry... » 2005-10-23 12:49:35

seiya_001
Replies: 4

umm..can anybody tell me some line (or is it curve..) equations? you know stuff like y=2x is linear equation.. smile i know it's lame but i really need it.. thanks

#8 Re: Help Me ! » Math Study Tips... » 2005-10-09 14:53:46

seiya_001

bro what you need in math is actually simple: PRACTICE...!! unlike other subject thatwe can understand by just reading them, math is something that YOU have to DO by yourself... only then you can understand it... and trust me, you're not alone... smile

#9 Re: Help Me ! » differential equation... (i think..) » 2005-10-05 12:36:24

seiya_001

thanks a lot dude..i'll try to find it in my country.. i'm sure we have it somewhere...:)

#10 Re: Help Me ! » differential equation... (i think..) » 2005-09-26 19:26:02

seiya_001
kybasche wrote:

I don't know if you have access to a Diff.Eq. book, but there should be a list of the different things that you can end up with depending on the right hand side.

any good reference? im' in indonesia so it's rather hard to get a good book with the original language (usually they're translated into indonesian and sometimes it's confusing), but shoot anyway...? smile

#11 Re: Help Me ! » differential equation... (i think..) » 2005-09-26 15:04:25

seiya_001
kybasche wrote:

Have you solved diffeq's before?  If not, I can see how it would seem a bit daunting smile

yes, but not with high order equations and all (d^2q/dt^2 or higher)

a question, what if the right hand side of the equation is not constant, let say it's a function also then what would you do with the 'particular solution'? thanks before..(btw in what year are you in college?)

#12 Re: Help Me ! » differential equation... (i think..) » 2005-09-25 17:28:31

seiya_001

wow, how do you get all that..!!
thanks a lot dude..i'll need some time to understand it all but i'll try...:)

#13 Re: Help Me ! » differential equation... (i think..) » 2005-09-25 13:57:31

seiya_001
kybasche wrote:

in my opinion, the easiest way to solve this would be to use laplace transforms.  do you know how to use them?

umm..not yet.., what about the other method you mentioned? can it be solved by any diffferential equation? thanx..:)

#14 Help Me ! » differential equation... (i think..) » 2005-09-19 22:08:23

seiya_001
Replies: 10

help....
how do you find the Q from:

d^2q        dq
------ + 8 ----- + 25 q = 300
dt^2         dt


any help would be MOST appreciated...

#15 Re: Help Me ! » Integration by substitution » 2005-09-13 18:31:22

seiya_001

I hope this helps...
your problem is:
∫2x^2/(2x^2+1) dx

divide 2x^2/(2x^2+1) to acquire 1-(1/2x^2+1)

then: ∫ 1-(1/2x^2+1) dx = ∫dx - ∫(1/2x^2+1) dx, yay..an easier integration...!!
                                    = x-∫(1/2x^2+1) dx, and we know that ∫1/(u^2+a^2) dx = 1/a arc tg u/a
thus: x- (arc tg x√2) or x(1-arc tg √2) + C

guys please CMIIW...

#16 Re: Help Me ! » Intergration Rock Solid!!! Please Help Asap.. » 2005-08-28 14:56:47

seiya_001
boombastictiger wrote:

4. A curve has the following equation: y=x³ -12x+7

(a)
(i) Find the gradient function ,dy?dx, of the curve.   (2 marks)

(ii) Find the coordinates of the points on the curve at which the gradient is zero.  (6 marks)

(iii) Describe the nature of the curves turning points.  (4 marks)

(
Challenging....

if y=x^3-12x+7
then
i) dy/dx = 3x^2 - 12

ii)dy/dx=0, 0=3x^2-12, 3x^2=12
x=√(12/3)=√4=2
y=2^3 - 12.2 +7
= 8-24+7=-9

iii) i don't understand the question, sorry..

guys, please correct me if i'm wrong...

#17 Re: Introductions » hiii... » 2005-08-28 13:49:03

seiya_001

well, it's a nice place. I live in Jakarta City. it' s the capital of indonesia. the people are very friendly here. maybe you should come here on vacation flowers4carlos.. smile

#18 Introductions » hiii... » 2005-08-25 14:55:15

seiya_001
Replies: 7

hi guys, i just find out about this awesome forum.. I'm a freshmen in one of the colleges in Indonesia and i love to learn about math... smile

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